Calculation of Punch Strength

In my exploration of metalworking and fabrication, I often encounter the critical aspect of Punch Strength Calculation. Understanding how to accurately calculate the punch strength is essential for ensuring the efficiency and safety of operations. By determining the appropriate strength required for various applications, we can optimize performance, reduce material wastage, and enhance the longevity of our tools. In this article, I will share insights into the methods and factors influencing punch strength, helping you to make informed decisions in your projects. Let’s dive into the calculations that underpin this vital process.

There are cases where trouble, such as punch tip breakage and flange fractures, occurs during the punching operation.

Often the cause of this trouble is a lack of technical data concerning standard parts, or an error in the selection of the punching tool material or shape. In order to reduce the incidence of this kind of trouble, standards for correct punch use, with consideration for factors such as the fatigue strength of tool steel and concentration of stress at flanges, are presented here.

1. Calculation of Punch Strength

● Punching force P［kgf］

P＝ ℓtτ… ………（1）ℓ : Punching profile length［mm］（For a round punch, ℓ＝πd）t : Material thickness［mm］

τ : Material shearing resistance［kgf/mm2］（τ≒0.8XTensile strengthσB）

［Example 1］ The maximum punching strength P when punching a round hole of diameter 2.8 mm in a high-tensile steel sheet of thickness 1.2 mm（tensile strength 80 kgf/mm2）, is the following. When P＝ℓtτ, Shearing resistanceτ ＝0.8×80＝64［kgf/mm2］

P＝3.14×2.8×1.2×64＝675 kgf

2. Fracture of punch tip

● Stress applied to punch tipσ［kgf/mm2］

σ＝P/A  P : Punching force, A : Cross-section area of punch tip（a ）For shoulder punch

σs＝4 tτ/d… ……………………（2） （b ）For jector punch σJ＝4d tτ/（d2－d12）………………（3）

［Example 2］ Find the possibility of punch tip fracture when shoulder punch SPAS6－50－P2.8 and Jector punch SJAS6－50－P2.8（d1 dimension＝0.7, as shown on P. 186）are used（. Punching conditions are the same as in Example 1.）

（a）For the shoulder punch, from Formula（2）: σs＝4×1.2×64/2.8＝110 kgf/mm2

（b）For the jector punch, from Formula（3）:σJ＝4×2.8×1.2×64 /（2.82－0.72）＝117 kgf/mm2

From Fig. 2, we see that whenσs is 110 kgf/mm2, there is the possibility of fracture occurring with an D2 punch at approximately 9,000 shots.

When the material is changed to M2, this increases to approximately 40,000 shots. The possibility for the jector punch is found in the same way.

Because the cross-section area is smaller, the punch tip will fracture at approximately 5,000 shots. Fracture will not occur if the stress applied to the punch during use is less than the maximum allowable stress for that punch material.（Consider this to be only a guide however, because the actual value varies depending on variations in the the die accuracy, die structure, and punched material, as well as the surface roughness, heat treatment, and other conditions of the punch.）

3. Minimum punching diameter

● Minimum punching diameter: dmin.  dmin＝4tτ/σ  σ: Fatigue strength of tool steel［kgf/mm2］

［Example 3］ The minimum punching diameter that is possible  when punching 100,000 shots or more in SPCC of thickness 2 mm with an M2 punch is the following. dmin ＝4tτ/σ……………（4） ＝4×2×26/97≒2.1mm Fatigue strength for M2 at 100,000

shots:σ＝97 kgf/mm2（from Fig. 2）τ ＝26 kgf/mm2（from Table 1）

4. Fracture due to buckling

● Buckling load P［kgf］ P＝nπ2EI/ℓ2 ………………（5） ℓ＝√ nπ2EI/P ………………（6） n : Coefficient n＝1 : Without stripper guide

n＝2 : With stripper guide  I : Second moment of inerti［a mm4］ For a round punch, I＝πd4/64  ℓ : Punch tip length［mm］

E : Young´s modulus［kgf/mm2］ D2 : 21000 M2 : 22000 HAP40 : 23000 V30 : 56000

As indicated by Euler´s formula, steps which can be take to improve buckling strength P include the use of a stripper guide, the use of a material with a larger Young’s modulus（SKD→SKH→HAP）, and reducing the punch tip length. The buckling load P indicates the load at the time when a punch buckles and fractures. When selecting a punch, it is therefore necessary to consider a safety factor of 3～5. When selecting a punch for punching small holes, special attention must be paid to the buckling load and to the stress which is applied to the punch.

［Example 4］Calculate the full length of the punch which will not produce buckling when aφ8 hole is punched in stainless steel 304（sheet thickness 1 mm, tensile strengthσb ＝60 kgf/mm2）with a straight punch（D2）. From Formula（6）: ℓ ＝√ nπ2EI/P＝√ 2×π2×21000× 201/1206＝262 mm If the safety factor is 3, then ℓ＝262/3＝87 mm If the punch plate sheet thickness t is 20 mm, then buckling can be prevented by using a punch of total length 107 mm or less. For a punch based on the stripper plate（punch plate tip is guided by the clearance）, the full length should be 87 mm or less.

［Example 5］ The buckling load P when a SHAL5－60－P2.00－BC20 punch is used without a stripper guide is the following.

P ＝nπ2EI/ℓ2＝1×π2×22000×0.785/202＝426 kgf

  If the safety factor is 3, then P＝426/3＝142 kgf Buckling will not occur at a punching force of 142 kgf or less.